Question

pls give me answer I WL Mark you brainliest​

Answer 1

Answer:

the coordinates of A are 0,-10

B are 4,0

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Answer 2

$\sf\huge\bold{Answer:}$

Given :

Line intersect x-axis at A

and y-axis at C.

Midpoint of line segment AC = B (-5,9)

To find :

Distance of A from O origin. i.e. OA.

Solution :

A will be of type (x, 0)

and C will be of type (0,y)

Mid point formula :

$\boxed{ \tt\red{ mid \: point = \bigg(\frac{x_1 + x_2 }{2} , \frac{y_1 +y_2}{2} }} \bigg)$

$\tt x_1$ = x coordinate of A = x

$\tt x_2$ = x coordinate of C = 0

$\tt y_1$ = y coordinate of A = 0

$\tt y_2$ = y coordinate of C = y

$\tt\ ( - 5,9) = \bigg( \frac{x + 0}{2}, \frac{0 + y}{2} \bigg)$

$\tt\ ( - 5,9) = \bigg(\frac{x}{2} , \frac{y}{2} \bigg)$

• Comparing x and y axis on both sides, we get

$\tt\ :⟶ - 5 = \frac{x}{2} and \: \tt\ 9 = \frac{y}{2}$

$\tt\ :⟶ \: x = - 5 \times 2,y = 9 \times 2$

$\tt\ :⟶ \: x = - 10,y = 18$

x = -10

y = 18

• A (x, 0) = (-10,0)
• C(0,y) = (0,18)

Distance formula :

$\boxed{\tt \red{d = \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} } } }$

point A (-10,0) Origin O (0,0)

$\tt x_1$ = x coordinate of A = -10

$\tt x_2$ = x coordinate of O = 0

$\tt y_1$ = y coordinate of A = 0

$\tt y_2$ = y coordinate of C = 0

$\tt\ OA = \sqrt{(0 - ( - 10)) {}^{2} + (0 - 0) {}^{2} }$

$\tt\ OA = \sqrt{(10) {}^{2} + (0) {}^{2} }$

$\tt\ OA = \sqrt{100 + 0}$

$\tt\ OA = \sqrt{10 \times 10}$

$\tt\ OA = 10 \: units</p><p>$

Distance of A from O origin :-

$\huge\tt \underline\purple{ 10 \: units}$